Discovering Liouville Integrability in Classical Mechanics

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The discussion focuses on demonstrating that a classical particle in a radial potential is Liouville integrable by analyzing its Hamiltonian in polar coordinates. The Hamiltonian is expressed in terms of generalized coordinates and velocities, prompting a question about the computation of the Poisson bracket with respect to these variables. It is clarified that in Hamiltonian dynamics, generalized velocities must be replaced with canonical momenta for proper formalism. The Lagrangian is provided, leading to the identification of two first integrals: the Hamiltonian itself, which is conserved due to time independence, and the canonical momentum associated with the cyclic variable φ, confirming the system's integrability. The discussion concludes with the affirmation that the system has two integrals of motion, making it integrable.
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Hi

Let a classical particle with unit mass subjected to a radial potential V and moving in a plane.

The Hamiltonian is written using polar coordinates (r,\phi)

H(r,\phi) = \frac{1}{2}(\dot{r}^2+r^2\dot{\phi}^2) - V(r)

I consider the angular momentum modulus C=r^2\dot{\phi},
and I want to show that the system is Liouville integrable (the problem is planar so I have to find a first integral (that is ,C) which commute to the Hamiltonian).

My question is : when I want to compute the Poisson bracket {H,C}, the only variable is r and \phi ? Because the conjuguate variables \dot{r} and \dot{\phi} appeared in H...
So I should write H(r,\phi,\dot{r},\dot{\phi}) and compute the partial derivative with respect to these four variables ?

Thanks
 
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somes explanations ?
 
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In Hamiltonian dynamics you must substitute the generalized velocities with the canonical momenta. Otherwise the formalism doesn't work properly. There's an alternative formalism, where you only perform the Legendre transformation from the Lagrange to the Hamilton formalisms for some generalized coordinates, which goes under the name of Routh, but I've never seen an application of it in practice (if you are interested in that, see A. Sommerfeld, Lectures on Theoretical Physics, Vol. 1).

In your case the Lagrangian should read
L=\frac{m}{2} (\dot{r}^2+r^2 \dot{\phi}^2) - V(r).
The canonically conjugated momenta thus are
p_r=\frac{\partial L}{\partial \dot{r}}=m \dot{r}, \quad p_{\phi}=\frac{\partial L}{\partial \dot{\phi}}=m r^2 \dot{\phi}.
The Hamiltonian now is
H(q,p)=\dot{q} \cdot p-L=\frac{m}{2} (\dot{r}^2+r^2 \dot{\phi}^2) +V(r) = \frac{1}{2m} \left (p_r^2 +\frac{p_{\phi}^2}{r^2} \right ) + V(r).
Now you have two first integrals:

(a) the Hamiltonian doesn't depend explicitly on time. Thus it is conserved:
H=E=\text{const}.
(b) the variable \phi is cyclic, i.e., the Hamiltonian doesn't depend on it. Thus the canonical momentum conjugate to this variable is conserved too:
p_{\phi}=\text{const}.
Thus you have two integral of motion for a system with two degrees of freedom. It is thus integrable.
 
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